What is the integral of (x^2-3x+11)/((x^2-4x+10)^2) ?

The integral of (x^2-3x+11)/((x^2-4x+10)^2) is

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integral of (x^2-3x+11)/((x^2-4x+10)^2)

\int \frac{x ^{2} - 3 x + 11}{( x ^{2} - 4 x + 10 ) ^{2}} d x

\frac{1}{12 6} x arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} x sin (2 arctan (\frac{1}{6} (x - 2))) + \frac{13}{12 6} arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} sin (2 arctan (\frac{1}{6} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{12 6 ( x ^{2} - 4 x + 10 )} - \frac{1}{2 ( x ^{2} - 4 x + 10 )} + C

Solution steps

\int \frac{x ^{2} - 3 x + 11}{( x ^{2} - 4 x + 10 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 3 x + 11}{( x ^{2} - 4 x + 10 ) ^{2}} : \frac{1}{x ^{2} - 4 x + 10} + \frac{x + 1}{( x ^{2} - 4 x + 10 ) ^{2}}

= \int \frac{1}{x ^{2} - 4 x + 10} + \frac{x + 1}{( x ^{2} - 4 x + 10 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{2} - 4 x + 10} d x + \int \frac{x + 1}{( x ^{2} - 4 x + 10 ) ^{2}} d x

\int \frac{1}{x ^{2} - 4 x + 10} d x = \frac{1}{6} arctan (\frac{x - 2}{6})

\int \frac{x + 1}{( x ^{2} - 4 x + 10 ) ^{2}} d x = \frac{1}{24 6} (2 arctan (\frac{x - 2}{6}) + sin (2 arctan (\frac{x - 2}{6}))) (x + 1) - \frac{1}{12} \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{6 ( x ^{2} - 4 x + 10 )} + \frac{6}{x ^{2} - 4 x + 10}

= \frac{1}{6} arctan (\frac{x - 2}{6}) + \frac{1}{24 6} (2 arctan (\frac{x - 2}{6}) + sin (2 arctan (\frac{x - 2}{6}))) (x + 1) - \frac{1}{12} \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{6 ( x ^{2} - 4 x + 10 )} + \frac{6}{x ^{2} - 4 x + 10}

Simplify

\frac{1}{6} arctan (\frac{x - 2}{6}) + \frac{1}{24 6} (2 arctan (\frac{x - 2}{6}) + sin (2 arctan (\frac{x - 2}{6}))) (x + 1) - \frac{1}{12} \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{6 ( x ^{2} - 4 x + 10 )} + \frac{6}{x ^{2} - 4 x + 10} : \frac{1}{12 6} x arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} x sin (2 arctan (\frac{1}{6} (x - 2))) + \frac{13}{12 6} arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} sin (2 arctan (\frac{1}{6} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{12 6 ( x ^{2} - 4 x + 10 )} - \frac{1}{2 ( x ^{2} - 4 x + 10 )}

= \frac{1}{12 6} x arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} x sin (2 arctan (\frac{1}{6} (x - 2))) + \frac{13}{12 6} arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} sin (2 arctan (\frac{1}{6} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{12 6 ( x ^{2} - 4 x + 10 )} - \frac{1}{2 ( x ^{2} - 4 x + 10 )}

Add a constant to the solution

= \frac{1}{12 6} x arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} x sin (2 arctan (\frac{1}{6} (x - 2))) + \frac{13}{12 6} arctan (\frac{1}{6} (x - 2)) + \frac{1}{24 6} sin (2 arctan (\frac{1}{6} (x - 2))) - \frac{( x - 2 ) ( arctan ( \frac{1}{6} ( x - 2 ) ) ( x ^{2} - 4 x + 10 ) + 6 ( x - 2 ) )}{12 6 ( x ^{2} - 4 x + 10 )} - \frac{1}{2 ( x ^{2} - 4 x + 10 )} + C

f (x) = \frac{x ^{3}}{4 - x ^{2}}

\int 1 + u^{2} d u

\int_{0}^{4} x^{3} x^{2} + 16 d x

\frac{d}{d x} (3 x^{2} + 2 x + 1)

t^{2} y^{^{'}} + t y = 5

What is the integral of (x^2-3x+11)/((x^2-4x+10)^2) ?
The integral of (x^2-3x+11)/((x^2-4x+10)^2) is